Solve a^2-18a-40=

$$a ^ { 2 } - 18 a - 40 =$$

Factor

$$\left(a-20\right)\left(a+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-40. To find p and q, set up a system to be solved.

$$p+q=-18 pq=1\left(-40\right)=-40$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -18.

$$p=-20 q=2$$

Rewrite a^{2}-18a-40 as \left(a^{2}-20a\right)+\left(2a-40\right).

$$\left(a^{2}-20a\right)+\left(2a-40\right)$$

Factor out a in the first and 2 in the second group.

$$a\left(a-20\right)+2\left(a-20\right)$$

Factor out common term a-20 by using distributive property.

$$\left(a-20\right)\left(a+2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$a^{2}-18a-40=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$a=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-40\right)}}{2}$$

Square -18.

$$a=\frac{-\left(-18\right)±\sqrt{324-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$a=\frac{-\left(-18\right)±\sqrt{324+160}}{2}$$

Add 324 to 160.

$$a=\frac{-\left(-18\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$a=\frac{-\left(-18\right)±22}{2}$$

The opposite of -18 is 18.

$$a=\frac{18±22}{2}$$

Now solve the equation a=\frac{18±22}{2} when ± is plus. Add 18 to 22.

$$a=\frac{40}{2}$$

Divide 40 by 2.

$$a=20$$

Now solve the equation a=\frac{18±22}{2} when ± is minus. Subtract 22 from 18.

$$a=-\frac{4}{2}$$

Divide -4 by 2.

$$a=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and -2 for x_{2}.

$$a^{2}-18a-40=\left(a-20\right)\left(a-\left(-2\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$a^{2}-18a-40=\left(a-20\right)\left(a+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -18x -40 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 18 rs = -40$$

Two numbers r and s sum up to 18 exactly when the average of the two numbers is \frac{1}{2}*18 = 9. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 9 - u s = 9 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -40

$$(9 - u) (9 + u) = -40$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$81 - u^2 = -40$$

Simplify the expression by subtracting 81 on both sides

$$-u^2 = -40-81 = -121$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 121 u = \pm\sqrt{121} = \pm 11$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =9 - 11 = -2 s = 9 + 11 = 20$$

Evaluate

$$\left(a-20\right)\left(a+2\right)$$